Answer
$-9x+y+7z=4$
Work Step by Step
The standard equation of a plane passing through the point $(x_1,y_1,z_1)$ can be defined as:
$a(x-x_1)+b(y-y_1)+c(z-z_1)=0$
The normal vector to the plane is given by:$n=\lt -9,1,7 \gt$
Thus, $-9(x -1)+1(y +1)+7(z-2)=0 \implies -9x+9+y+1+7z-14=0$
or, $-9x+y+7z=4$
